The value of the definite integral ∫0π/2sin|2x−α|dx where α∈[0,π]

I=∫−α(x−αsin|t|dt where 2x−α=t⇒dx=dt2=12∫−α0−sint dt+12∫0 π−αsin t dt=[12cos t]−α0+[−12cot]0 π−α=12[1−cos α]−12[−cosα−1]=12(1−cosα)+12(1+cosα)=1

The value of the definite integral ∫0π/2sin|2x−α|dx where α∈[0,π]

I=∫−α(x−αsin|t|dt where 2x−α=t⇒dx=dt2=12∫−α0−sint dt+12∫0 π−αsin t dt=[12cos t]−α0+[−12cot]0 π−α=12[1−cos α]−12[−cosα−1]=12(1−cosα)+12(1+cosα)=1

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The value of the definite integral $\int_{0}^{π / 2} \sin | 2 x - α | d x$ where $α \in [0, π]$

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$\cos α$

$\frac{1 + \cos α}{2}$

$\frac{1 - \cos α}{2}$

answer is A.

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Detailed Solution

$\begin{array}{l} I = \int_{- α}^{(x - α} \sin | t | d t w h e r e 2 x - α = t \\ \Rightarrow d x = \frac{d t}{2} \\ = \frac{1}{2} \int_{- α}^{0} - \sin t d t + \frac{1}{2} \int_{0}^{^{π - α}} \sin t d t \\ = {[\frac{1}{2} \cos t]}_{- α}^{0} + {[- \frac{1}{2} \cot]}_{0^{}}^{π - α} \\ = \frac{1}{2} [1 - \cos α] - \frac{1}{2} [- \cos α - 1] \\ = \frac{1}{2} (1 - \cos α) + \frac{1}{2} (1 + \cos α) = 1 \end{array}$